Question

You intend to estimate a population mean mu with the following sample.
87.1
69.8
94.3
94.7
76.3
88.4
95.3
87.8
109.5
85.8
102.5
85.6


You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

80% C.I. =

Answer 1

To calculate the 80% confidence interval for the population mean \(\mu\), we will follow these steps:

1. **Calculate the Sample Mean (\(\bar{x}\)) and Sample Standard Deviation (s)**.
2. **Find the critical value (t*) for 80% confidence level**.
3. **Calculate the Margin of Error**.
4. **Determine the Confidence Interval**.

### Step 1: Calculate the Sample Mean and Sample Standard Deviation

Given the sample values:
- 87.1
- 69.8
- 94.3
- 94.7
- 76.3
- 88.4
- 95.3
- 87.8
- 109.5
- 85.8
- 102.5
- 85.6

#### Calculate the sample mean (\(\bar{x}\)):
\[
\bar{x} = \frac{\sum x_i}{n} = \frac{87.1 + 69.8 + 94.3 + 94.7 + 76.3 + 88.4 + 95.3 + 87.8 + 109.5 + 85.8 + 102.5 + 85.6}{12} = \frac{ 1,030.0 }{12} \approx 85.83
\]

#### Calculate the sample standard deviation (s):
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]

Where:
- \(x_i\) are the sample values,
- \(\bar{x}\) is the sample mean,
- \(n\) is the number of samples.

Calculating the variance:
- First calculate \( (x_i - \bar{x})^2 \):

\[
\begin{align*}
(87.1 - 85.83)^2 & \approx 1.6349 \\
(69.8 - 85.83)^2 & \approx 258.0369 \\
(94.3 - 85.83)^2 & \approx 70.0039 \\
(94.7 - 85.83)^2 & \approx 78.6249 \\
(76.3 - 85.83)^2 & \approx 90.7689 \\
(88.4 - 85.83)^2 & \approx 6.5049 \\
(95.3 - 85.83)^2 & \approx 87.4569 \\
(87.8 - 85.83)^2 & \approx 3.8469 \\
(109.5 - 85.83)^2 & \approx 566.5449 \\
(85.8 - 85.83)^2 & \approx 0.0009 \\
(102.5 - 85.83)^2 & \approx 284.7849 \\
(85.6 - 85.83)^2 & \approx 0.0529 \\
\end{align*}
\]

The sum of these squared differences is:
\[
1.6349 + 258.0369 + 70.0039 + 78.6249 + 90.7689 + 6.5049 + 87.4569 + 3.8469 + 566.5449 + 0.0009 + 284.7849 + 0.0529 \approx 1,470.40
\]

Calculating the sample variance:
\[
s^2 = \frac{1,470.40}{12 - 1} \approx 133.67 \quad \Rightarrow \quad s \approx 11.56
\]

### Step 2: Find the Critical Value (t*) for 80% Confidence Level

For an 80% confidence level and \(n - 1 = 11\) degrees of freedom, we refer to the t-distribution table:

\[
t^* \approx 1.363 (approx)
\]

### Step 3: Calculate the Margin of Error

The margin of error \(E\) is calculated using the formula:
\[
E = t^* \frac{s}{\sqrt{n}} = 1.363 \cdot \frac{11.56}{\sqrt{12}} \approx 1.363 \cdot 3.344 \approx 4.56
\]

### Step 4: Determine the Confidence Interval

The 80% confidence interval is:
\[
(\bar{x} - E, \bar{x} + E) = (85.83 - 4.56, 85.83 + 4.56) \approx (81.27, 90.39)
\]

Therefore, the 80% confidence interval for the population mean \(\mu\) is:
\[
\text{80% C.I.} = (81.27, 90.39)
\]

So, the final answer is:
\[
(81.27, 90.39)
\]